What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x + 5} = \dfrac{-6x - 6}{x + 5}$
Answer: Multiply both sides by $x + 5$ $ \dfrac{x^2 - 1}{x + 5} (x + 5) = \dfrac{-6x - 6}{x + 5} (x + 5)$ $ x^2 - 1 = -6x - 6$ Subtract $-6x - 6$ from both sides: $ x^2 - 1 - (-6x - 6) = -6x - 6 - (-6x - 6)$ $ x^2 - 1 + 6x + 6 = 0$ $ x^2 + 5 + 6x = 0$ Factor the expression: $ (x + 1)(x + 5) = 0$ Therefore $x = -1$ or $x = -5$ At $x = -5$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -5$, it is an extraneous solution.